Binding a const function reference to a lambdaWhat is a lambda (function)?What are the differences between a pointer variable and a reference variable in C++?Why are Python lambdas useful?What is the difference between const int*, const int * const, and int const *?How come a non-const reference cannot bind to a temporary object?C++0x lambda capture by value always const?Returning unique_ptr from functionsWhat is a lambda expression in C++11?Returning temporary object and binding to const referenceReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviations

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Binding a const function reference to a lambda


What is a lambda (function)?What are the differences between a pointer variable and a reference variable in C++?Why are Python lambdas useful?What is the difference between const int*, const int * const, and int const *?How come a non-const reference cannot bind to a temporary object?C++0x lambda capture by value always const?Returning unique_ptr from functionsWhat is a lambda expression in C++11?Returning temporary object and binding to const referenceReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviations






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








16















This code



int(&foo)(int, int) = [](int a, int b) return a + b; ;


doesn't compile since apparently a non-const reference can't be initialized with a temporary. Where do I put the const?










share|improve this question
























  • Why does it need to be a const ref? By making a copy the compiler will make use of the move semantics

    – Narase
    Jul 1 at 4:41











  • It doesn't compile because a lambda does not have conversion to reference. The compiler error is misleading.

    – cpplearner
    Jul 1 at 4:43






  • 1





    Why not just use auto foo = [](int a, int b) return a + b; ;? C++ will move instead of copy the temporary.

    – Yang
    Jul 1 at 4:45












  • It doesn't need to be, I'm just curious how/if this can be done.

    – Artikash
    Jul 1 at 4:58

















16















This code



int(&foo)(int, int) = [](int a, int b) return a + b; ;


doesn't compile since apparently a non-const reference can't be initialized with a temporary. Where do I put the const?










share|improve this question
























  • Why does it need to be a const ref? By making a copy the compiler will make use of the move semantics

    – Narase
    Jul 1 at 4:41











  • It doesn't compile because a lambda does not have conversion to reference. The compiler error is misleading.

    – cpplearner
    Jul 1 at 4:43






  • 1





    Why not just use auto foo = [](int a, int b) return a + b; ;? C++ will move instead of copy the temporary.

    – Yang
    Jul 1 at 4:45












  • It doesn't need to be, I'm just curious how/if this can be done.

    – Artikash
    Jul 1 at 4:58













16












16








16


3






This code



int(&foo)(int, int) = [](int a, int b) return a + b; ;


doesn't compile since apparently a non-const reference can't be initialized with a temporary. Where do I put the const?










share|improve this question
















This code



int(&foo)(int, int) = [](int a, int b) return a + b; ;


doesn't compile since apparently a non-const reference can't be initialized with a temporary. Where do I put the const?







c++ c++11 lambda






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jul 1 at 8:21









StoryTeller

114k18 gold badges245 silver badges311 bronze badges




114k18 gold badges245 silver badges311 bronze badges










asked Jul 1 at 4:33









ArtikashArtikash

3153 silver badges11 bronze badges




3153 silver badges11 bronze badges












  • Why does it need to be a const ref? By making a copy the compiler will make use of the move semantics

    – Narase
    Jul 1 at 4:41











  • It doesn't compile because a lambda does not have conversion to reference. The compiler error is misleading.

    – cpplearner
    Jul 1 at 4:43






  • 1





    Why not just use auto foo = [](int a, int b) return a + b; ;? C++ will move instead of copy the temporary.

    – Yang
    Jul 1 at 4:45












  • It doesn't need to be, I'm just curious how/if this can be done.

    – Artikash
    Jul 1 at 4:58

















  • Why does it need to be a const ref? By making a copy the compiler will make use of the move semantics

    – Narase
    Jul 1 at 4:41











  • It doesn't compile because a lambda does not have conversion to reference. The compiler error is misleading.

    – cpplearner
    Jul 1 at 4:43






  • 1





    Why not just use auto foo = [](int a, int b) return a + b; ;? C++ will move instead of copy the temporary.

    – Yang
    Jul 1 at 4:45












  • It doesn't need to be, I'm just curious how/if this can be done.

    – Artikash
    Jul 1 at 4:58
















Why does it need to be a const ref? By making a copy the compiler will make use of the move semantics

– Narase
Jul 1 at 4:41





Why does it need to be a const ref? By making a copy the compiler will make use of the move semantics

– Narase
Jul 1 at 4:41













It doesn't compile because a lambda does not have conversion to reference. The compiler error is misleading.

– cpplearner
Jul 1 at 4:43





It doesn't compile because a lambda does not have conversion to reference. The compiler error is misleading.

– cpplearner
Jul 1 at 4:43




1




1





Why not just use auto foo = [](int a, int b) return a + b; ;? C++ will move instead of copy the temporary.

– Yang
Jul 1 at 4:45






Why not just use auto foo = [](int a, int b) return a + b; ;? C++ will move instead of copy the temporary.

– Yang
Jul 1 at 4:45














It doesn't need to be, I'm just curious how/if this can be done.

– Artikash
Jul 1 at 4:58





It doesn't need to be, I'm just curious how/if this can be done.

– Artikash
Jul 1 at 4:58












2 Answers
2






active

oldest

votes


















24














As mentioned already, a capture-less lambda is convertible to a function pointer. So if you want to bind that static function to a reference, you need to dereference the pointer.



int(&foo)(int, int) = *[](int a, int b) return a + b; ;


Applying * to the lambda causes a bunch of machinery to kick in. Since the lambda doesn't overload operator*, but does implement a conversion to a pointer type, that conversion happens. Afterwards * is applied to the returned pointer and that yields a function lvalue. That lvalue can then bind to the reference.



Here it is live.






share|improve this answer






























    11














    A lambda can only be converted to a function pointer if it does not capture.




    The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator




    [Lambda Functions][1]


    I changed your code as below and it worked.



    int (*foo)(int, int)= [] (int a, int b) return a + b; ;
    int main()

    cout << "Res:: " << foo(10,20);
    return 0;



    I just make it function pointer.



    Alternatively,



    auto foo = [](int a, int b) return a + b; ; 


    is also a good choice.



    I hope it helps!






    share|improve this answer

























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      24














      As mentioned already, a capture-less lambda is convertible to a function pointer. So if you want to bind that static function to a reference, you need to dereference the pointer.



      int(&foo)(int, int) = *[](int a, int b) return a + b; ;


      Applying * to the lambda causes a bunch of machinery to kick in. Since the lambda doesn't overload operator*, but does implement a conversion to a pointer type, that conversion happens. Afterwards * is applied to the returned pointer and that yields a function lvalue. That lvalue can then bind to the reference.



      Here it is live.






      share|improve this answer



























        24














        As mentioned already, a capture-less lambda is convertible to a function pointer. So if you want to bind that static function to a reference, you need to dereference the pointer.



        int(&foo)(int, int) = *[](int a, int b) return a + b; ;


        Applying * to the lambda causes a bunch of machinery to kick in. Since the lambda doesn't overload operator*, but does implement a conversion to a pointer type, that conversion happens. Afterwards * is applied to the returned pointer and that yields a function lvalue. That lvalue can then bind to the reference.



        Here it is live.






        share|improve this answer

























          24












          24








          24







          As mentioned already, a capture-less lambda is convertible to a function pointer. So if you want to bind that static function to a reference, you need to dereference the pointer.



          int(&foo)(int, int) = *[](int a, int b) return a + b; ;


          Applying * to the lambda causes a bunch of machinery to kick in. Since the lambda doesn't overload operator*, but does implement a conversion to a pointer type, that conversion happens. Afterwards * is applied to the returned pointer and that yields a function lvalue. That lvalue can then bind to the reference.



          Here it is live.






          share|improve this answer













          As mentioned already, a capture-less lambda is convertible to a function pointer. So if you want to bind that static function to a reference, you need to dereference the pointer.



          int(&foo)(int, int) = *[](int a, int b) return a + b; ;


          Applying * to the lambda causes a bunch of machinery to kick in. Since the lambda doesn't overload operator*, but does implement a conversion to a pointer type, that conversion happens. Afterwards * is applied to the returned pointer and that yields a function lvalue. That lvalue can then bind to the reference.



          Here it is live.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jul 1 at 5:26









          StoryTellerStoryTeller

          114k18 gold badges245 silver badges311 bronze badges




          114k18 gold badges245 silver badges311 bronze badges























              11














              A lambda can only be converted to a function pointer if it does not capture.




              The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator




              [Lambda Functions][1]


              I changed your code as below and it worked.



              int (*foo)(int, int)= [] (int a, int b) return a + b; ;
              int main()

              cout << "Res:: " << foo(10,20);
              return 0;



              I just make it function pointer.



              Alternatively,



              auto foo = [](int a, int b) return a + b; ; 


              is also a good choice.



              I hope it helps!






              share|improve this answer



























                11














                A lambda can only be converted to a function pointer if it does not capture.




                The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator




                [Lambda Functions][1]


                I changed your code as below and it worked.



                int (*foo)(int, int)= [] (int a, int b) return a + b; ;
                int main()

                cout << "Res:: " << foo(10,20);
                return 0;



                I just make it function pointer.



                Alternatively,



                auto foo = [](int a, int b) return a + b; ; 


                is also a good choice.



                I hope it helps!






                share|improve this answer

























                  11












                  11








                  11







                  A lambda can only be converted to a function pointer if it does not capture.




                  The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator




                  [Lambda Functions][1]


                  I changed your code as below and it worked.



                  int (*foo)(int, int)= [] (int a, int b) return a + b; ;
                  int main()

                  cout << "Res:: " << foo(10,20);
                  return 0;



                  I just make it function pointer.



                  Alternatively,



                  auto foo = [](int a, int b) return a + b; ; 


                  is also a good choice.



                  I hope it helps!






                  share|improve this answer













                  A lambda can only be converted to a function pointer if it does not capture.




                  The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator




                  [Lambda Functions][1]


                  I changed your code as below and it worked.



                  int (*foo)(int, int)= [] (int a, int b) return a + b; ;
                  int main()

                  cout << "Res:: " << foo(10,20);
                  return 0;



                  I just make it function pointer.



                  Alternatively,



                  auto foo = [](int a, int b) return a + b; ; 


                  is also a good choice.



                  I hope it helps!







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Jul 1 at 5:06









                  Abhishek SinhaAbhishek Sinha

                  3811 silver badge6 bronze badges




                  3811 silver badge6 bronze badges



























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